The minimum surface area that such a box can have is 380 square
<h3>How to determine the minimum surface area such a box can have?</h3>
Represent the base length with x and the bwith h.
So, the volume is
V = x^2h
This gives
x^2h = 500
Make h the subject
h = 500/x^2
The surface area is
S = 2(x^2 + 2xh)
Expand
S = 2x^2 + 4xh
Substitute h = 500/x^2
S = 2x^2 + 4x * 500/x^2
Evaluate
S = 2x^2 + 2000/x
Differentiate
S' = 4x - 2000/x^2
Set the equation to 0
4x - 2000/x^2 = 0
Multiply through by x^2
4x^3 - 2000 = 0
This gives
4x^3= 2000
Divide by 4
x^3 = 500
Take the cube root
x = 7.94
Substitute x = 7.94 in S = 2x^2 + 2000/x
S = 2 * 7.94^2 + 2000/7.94
Evaluate
S = 380
Hence, the minimum surface area that such a box can have is 380 square
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More than five lol sorry I am only trying to help
Answer:
The mean is 87 and the median is 87
Step-by-step explanation:
Answer:
2x²-32 ⇒ x²=16⇒ (-4,4)
4x²-100 ⇒x²=25 ⇒(-5,5)
x²-55=9 ⇒x²=64 ⇒(-8,8)
x²-140=-19 ⇒x²=121 ⇒(-11,11)
2x²-18=0 ⇒x²=9 ⇒(-3,3)
Answer:
10 i think
Step-by-step explanation:
